# Asymptotic Expansion to find leading behavior term

I am trying to find the asymptotic expansion to get the leading order term of $$I(x)=\int_0^{2\pi} (1+t^2) e^{x\cos t} dt$$ as $$x\to\infty$$.

The way I am approaching this is by first looking at the phase function $$\phi(t)=\cos t$$ which has two maxima in our interval of integration (one maxima at $$t=0$$ and one at $$t=2\pi$$). Since it has two maxima, I thought it might be a good idea to split the integral up into two integrals around each maxima. So, I write \begin{align} I(x)&=\int_0^{\pi} (1+t^2) e^{x\cos t} dt+\int_\pi^{2\pi} (1+t^2) e^{x\cos t} dt\\ &\sim\int_0^{\delta} (1+t^2) e^{x\cos t} dt+\int_{2\pi-\delta}^{2\pi} (1+t^2) e^{x\cos t} dt \end{align} since the major contribution is coming from the regions around the two maxima.

Just looking at the first integral, which I call $$I_A=\int_0^{\delta} (1+t^2) e^{x\cos t} dt$$, I can Taylor expand the phase function around $$t=0$$, so $$\phi(t)\approx 1-\frac{t^2}{2}+\frac{t^4}{24}-\ldots=1-\frac{t^2}{2}+\{HOT\}$$ where $$\{HOT\}$$ is higher order terms. For this to be justifiable, I need this Taylor Series to be ordered, so I need $$xt^4\ll \mathcal{O}(1)\implies x\delta^4\ll \mathcal{O}(1)\implies$$ we can pick $$\delta\ll x^{-1/4}\to 0$$ as $$x\to\infty$$.

Now, I can write $$I_A=e^x\int_0^\delta (1+t^2) e^{-\frac{xt^2}{2}} e^{x\{HOT\}} dt.$$

This looks similar to the form of Watson's lemma where I might be able to express this in terms of a Gamma function to get the asymptotic expansion, but I am not sure how to continue with Watson's lemma or if this is even correct since I have $$e^x$$ in front which is blowing up as $$x\to\infty$$, so it doesn't seem like a proper asymptotic expansion. Any help would be greatly appreciated.

• Does this help? Commented Sep 25, 2022 at 21:43
• @AaronHendrickson I thought of using Laplace's method but am not sure how exactly to apply it here? Commented Sep 25, 2022 at 21:54
• Did you look under "Other formulations" section? Your problem looks to be virtually in the exact form to directly apply. Agree? Commented Sep 25, 2022 at 21:55
• @AaronHendrickson Oh I see now, thanks for that! Would it still be a problem that I have $e^x$ multiplying outside my integral and that would blow up? Also, is what I am doing by splitting up the integral into two parts around each maxima valid? Commented Sep 25, 2022 at 21:58
• @AaronHendrickson And also, how to I treat the $e^{x\{HOT\}}$ terms that I have when I am applying Laplace method as you mentioned? Commented Sep 25, 2022 at 22:00

We have \begin{align*} I(x) = \int_0^{2\pi } {(1 + t^2 )\mathrm{e}^{x\cos t} \mathrm{d}t} & = \int_0^\pi {(1 + t^2 )\mathrm{e}^{x\cos t} \mathrm{d}t} + \int_\pi ^{2\pi } {(1 + t^2 )\mathrm{e}^{x\cos t} \mathrm{d}t} \\ & = \int_0^\pi {\left[ {2 + t^2 + (2\pi - t)^2 } \right]\mathrm{e}^{x\cos t} \mathrm{d}t} . \end{align*} Thus, by Laplace's method, $$I(x) = \sqrt {2\pi }(1 + 2\pi ^2 ) \mathrm{e}^x x^{ - 1/2} (1 + \mathcal{O}(x^{ - 1/2} ))$$ as $$x\to +\infty$$.